Tiêu đề 2. Low of motion (S.C.Q. ) E.pdf ✅

 μmg cos  = mg sin   tan  = μ = 2 1 ...(i) From equation y = 20 x 2  tan  = dx dy = 10 x .... (ii) From (i) & (ii) , x = 5 , so y = 1.25 Q.16 A particle moves on a rough horizontal ground with some initial velocity say v0. If 3/4th of its kinetic energy is lost in friction in time t0. Then coefficient of friction between the particle and the ground is- (A) 0 0 2gt v (B) 0 0 4gt v (C) 0 0 4gt 3v (D) 0 0 gt v Sol. [A] 4 3 th energy is lost i.e. 4 1 KE is left. Hence new velocity becomes 2 v0 under the retardation μg in time t0, 2 v0 = v0 – μg t0  μ = 0 0 2gt v Q.17 System shown in figure is released from rest. Pulley and spring is massless and friction is absent everywhere. The speed of 5 kg block when 2 kg block leaves contact with ground is : (Take force constant of spring k = 40 N/m and g = 10 m/s2 ) 5 kg 2 kg (A) 2 m/s (B) 2 2 m/s (C) 2 m/s (D) 4 2 m/s Sol. [B] Let x be the extension is spring when 2 kg block leaves the contact with ground. Then kx = 2g x = k 2g = 40 210 = m 2 1 Now from conservation of energy, mgx = 2 1 kx2 + 2 1 mv2  v = m kx 2gx – 2 = 2 2m/s Q.18 A system consists of two cubes of masses m1 and m2 respectively connected by a spring of force constant k. The force (F) that should be applied to the upper cube for which the lower one just lifts after the force is removed is- m2 k m1 (A) m1g (B) g m m m m 1 2 1 2  (C) (m1 + m2) g (D) m2g Sol. [C] Initially, F = m1g = k1x1 x1 = k F  m2g ....(1) Finally x2 = k m g2 .... (2) From conservation of energy, m1g(x1 + x2) = 2 1 k   2 2 2 x1 – x .... (3) From (1), (2) & (3), F = (m1 + m2)g Q.19 A block of mass m is placed at rest on an inclined plane of inclination  to the horizontal. If the coefficient of friction between the block and the plane is , then the total force the inclined plane exerts on the block is- (A) mg (B)  mg cos  (C) mg sin  (D)  mg tan  Sol. [A] mg cos  f = mg sin  mg sin  N In static condition, Friction, f = mgsin N = mg cos  so net force through plane, 2 2 (mgsin )  (mgcos) = mg Q.20 For the arrangement shown in figure the coefficient of friction between the two blocks is . If both the block are identical, then the acceleration of each block is- m m F (A) 2m F – 2g (B) 2m F (C) 2m F – g (D) zero Sol. [C] m m F T T f f In fig, direction of friction is shown,